Functions whose Fourier coefficients are all zero

Question

Suppose I have a measurable complex function on the circle $f : S^1 \rightarrow \mathbb{C}$ and I find its fourier coefficients $\int f(x) e^{-2 i n \pi x} dx$ are all 0.

Is $f$ a.e. 0? Could it be something else?

Edit: I figured out a proof that $f$ a.e. 0. But my proof only works on the circle. Is the same thing true for $f : \mathbb{R} \rightarrow \mathbb{C}$?

@DavidC.Ullrich Is it the same for $f : \mathbb{R} \rightarrow \mathbb{C}$? — Mark, Oct 24 '15 at 19:44
If $f\in L^1(\Bbb R)$ and $\hat f=0$ then $f=0$, yes. This is immediate from the Inversion Theorem. See any number of books... — David C. Ullrich, Oct 24 '15 at 20:55

score 0 · Accepted Answer · answered Oct 24 '15 at 19:33

0

Okay so we know $f$ is $L_1$ at least since the first coefficient is the integral. And we also know that for $L_1$ functions the arithmetic averages of partial sums of the fourier approximations converge in $L_1$ to the actual function. But the actual function must be 0 then.

answered Oct 24 '15 at 19:33

Mark

5,696

For $L_1$ functions the arithmetic averages of partial sums of the Fourier approximations converge in $L_1$ to the actual function. What's the name of this theorem? From Fejer's Theorem, we only know this holds at the continuous points of $f$. – U2647 Mar 10 '19 at 07:53

Functions whose Fourier coefficients are all zero

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